Is there any nonsolvable finite group $G$ such that ${\rm cd}(G) =\{1, 3, 4, 5, 6, 10 \}$, where ${\rm cd}(G)$ is the set of irreducible character degrees of $G$. In other words we have that a normal subgroup $M$ of $G$ such that $G/M \cong A_5$. Now, how do we can construct a nonsolvable finite group $G$ such that ${\rm cd}(G) =\{ 1, 3, 4, 5, a, b\}$ where $6$ divides $a$, $10$ divides $b$, and $(5, a)=(3,b)=1$ ?
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4$\begingroup$ I don't understand why the first sentence implies the second sentence $\endgroup$ – Yemon Choi Jul 22 '17 at 15:00

$\begingroup$ The first sentence is a special case of the second. In other words, the relation between the degrees are important. $\endgroup$ – Guest Jul 22 '17 at 15:40

2$\begingroup$ It is not true that ${\rm cd}(A_5) = \{1,2,3,4,5,6,10\}$. So the sentence beginning "In other words" is not true. $\endgroup$ – Derek Holt Jul 22 '17 at 16:08

1$\begingroup$ But then doesn't $G = 1^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 10^{2} = 187$ make that impossible, since all groups of order $187 = 11 \cdot 17$ are cyclic? $\endgroup$ – DavidLHarden Jul 22 '17 at 18:04

1$\begingroup$ @DavidLHarden this is the set of irreducible character degrees but my understanding is that it ignores multiplicities. $\endgroup$ – YCor Jul 22 '17 at 18:08